Why 1+12+ Y3 < 1100 
Says the state of university Need to purchase 1100 computers in total, we have the following answer on the way top
        
             
        
        
        
Answer:
(a) E = 0 N/C
(b) E = 0 N/C
(c) E = 7.78 x10^5 N/C
Explanation:
We are given a hollow sphere with following parameters:
Q = total charge on its surface = 23.6 μC = 23.6 x 10^-6 C
R = radius of sphere = 26.1 cm = 0.261 m
Permittivity of free space = ε0 = 8.85419 X 10−12 C²/Nm²
The formula for the electric field intensity is:
E = (1/4πεo)(Q/r²)
where, r = the distance from center of sphere where the intensity is to be found.
(a)
At the center of the sphere r = 0. Also, there is no charge inside the sphere to produce an electric field. Thus the electric field at center is zero.
<u>E = 0 N/C</u>
(b)
Since, the distance R/2 from center lies inside the sphere. Therefore, the intensity at that point will be zero, due to absence of charge inside the sphere (q = 0 C).
<u>E = 0 N/C</u>
(c)
Since, the distance of 52.2 cm is outside the circle. So, now we use the formula to calculate the Electric Field:
E = (1/4πεo)[(23.6 x 10^-6 C)/(0.522m)²]
<u>E = 7.78 x10^5 N/C</u>
 
        
             
        
        
        
Answer:
The process from a liquid to a vapor.
Explanation:
Water is evaporated by heating up and turning into a vapor. 
 
        
                    
             
        
        
        
Answer:
This should be true because program is a software and mouse and keyboard are specific computer hardware to make your exsperience with a computer more efficent and faster.
 
        
             
        
        
        
Answer:
The surface temperature of the ground is = 296.946K
Explanation:
Solution
Given
r₁= 0.05m
r₂= 0.08m
Tn =Ti = 77K
Ki = 0.0035 Wm-1K-1
Kg =  1 Wm-1K-1
Z= 2m
Now,
The outer type temperature (Skin temperature pipe)
Q = T₀ -T₁/ln (r2/r1)/2πKi = 2πKi  T0 -T1/ln (r2/r1)
Thus,
10 w/m = 2π * 0.0035 = T0 -77/ln 0.08/0.05
⇒ T₀ -77 = 231.72
     T₀= 290.72K
The shape factor between the cylinder and he ground
S = 2πL/ln 4z/D
where L = length of pipe
D = outer layer of pipe
S = 2π * 1/4 *2/ 2 * 0.08 = 1.606m
The heat gained in the pipe is = S  * Kg * (Tg- T₀)
(10* 1) = 1.606 * 1* (Tg- 290.72)
Tg - 290.72 = 6.2266
Tg = 296.946K
Therefore the surface temperature to the ground is 296.946K